我正在尝试使用lapply运行带有一堆输出的线性模型。我很难分辨哪个输出属于哪个参数对，所以我想在每个输出中粘贴列表名称。这里是一些数据：
Site Date Season Species BodySize Mass SeasonGDD
1 3/19/17 winter infrequens 6 1.134965577 3.556416667
1 3/19/17 winter infrequens 6 1.134965577 3.556416667
1 3/19/17 winter infrequens 6 1.134965577 3.556416667
1 3/19/17 winter infrequens 6 1.134965577 3.556416667
1 3/19/17 winter infrequens 6 1.134965577 3.556416667
1 3/19/17 winter infrequens 6.2 1.245407892 3.556416667
1 3/19/17 winter infrequens 6.2 1.245407892 3.556416667
1 3/19/17 winter infrequens 6.2 1.245407892 3.556416667
1 3/19/17 winter infrequens 6.4 1.362573803 3.556416667
1 3/19/17 winter infrequens 6.4 1.362573803 3.556416667
1 3/19/17 winter doddsii 6.4 1.362573803 3.556416667
1 3/19/17 winter infrequens 6.4 1.362573803 3.556416667
1 3/19/17 winter infrequens 6.4 1.362573803 3.556416667
4 3/19/17 winter infrequens 6.4 1.362573803 3.556416667
4 3/19/17 winter doddsii 6.6 1.486643287 3.556416667
3 3/19/17 winter infrequens 6.6 1.486643287 3.556416667
2 3/19/17 winter infrequens 6.6 1.486643287 3.556416667
2 3/19/17 winter infrequens 6.6 1.486643287 3.556416667
1 3/19/17 winter infrequens 6.6 1.486643287 3.556416667
1 3/19/17 winter infrequens 6.6 1.486643287 3.556416667
1 3/19/17 winter doddsii 6.6 1.486643287 3.556416667
1 3/19/17 winter doddsii 6.6 1.486643287 3.556416667
1 3/19/17 winter cockerelli 6.8 1.293245066 3.556416667
5 3/19/17 winter infrequens 6.8 1.617795375 3.556416667
5 3/19/17 winter infrequens 6.8 1.617795375 3.556416667
3 3/19/17 winter infrequens 6.8 1.617795375 3.556416667
1 3/19/17 winter infrequens 6.8 1.617795375 3.556416667
5 3/19/17 winter doddsii 6.8 1.617795375 3.556416667
4 3/19/17 winter grandis 6.9 1.347971145 3.556416667
1 3/19/17 winter doddsii 7 1.75620819 3.556416667
1 3/19/17 winter infrequens 7 1.75620819 3.556416667
1 3/19/17 winter infrequens 7 1.75620819 3.556416667
1 3/19/17 winter infrequens 7 1.75620819 3.556416667
1 3/19/17 winter doddsii 7.2 1.902058976 3.556416667
1 3/19/17 winter infrequens 7.2 1.902058976 3.556416667
2 3/19/17 winter infrequens 7.2 1.902058976 3.556416667
2 3/19/17 winter doddsii 7.4 2.055524128 3.556416667
2 3/19/17 winter doddsii 7.4 2.055524128 3.556416667
2 3/19/17 winter doddsii 7.4 2.055524128 3.556416667
1 3/19/17 winter doddsii 7.4 2.055524128 3.556416667
1 3/19/17 winter doddsii 7.4 2.055524128 3.556416667
1 3/19/17 winter cockerelli 7.5 1.707996411 3.556416667
1 3/19/17 winter doddsii 7.6 2.216779222 3.556416667
3 3/19/17 winter doddsii 7.6 2.216779222 3.556416667
4 3/19/17 winter doddsii 7.6 2.216779222 3.556416667
1 3/19/17 winter infrequens 7.6 2.216779222 3.556416667
1 3/19/17 winter cockerelli 7.7 1.840497347 3.556416667
1 3/19/17 winter doddsii 7.8 2.385999034 3.556416667
2 3/19/17 winter doddsii 7.8 2.385999034 3.556416667
2 3/19/17 winter doddsii 7.8 2.385999034 3.556416667
3 3/19/17 winter doddsii 8.2 2.749028097 3.556416667
1 3/19/17 winter doddsii 8.6 3.145994526 3.556416667
5 3/19/17 winter cockerelli 8.7 2.603054046 3.556416667
1 3/19/17 winter doddsii 9 3.578270231 3.556416667
5 3/19/17 winter doddsii 9.4 4.047216436 3.556416667
5 3/19/17 winter cockerelli 9.7 3.545136654 3.556416667
5 3/19/17 winter cockerelli 10.6 4.560698666 3.556416667
4 3/19/17 winter cockerelli 10.8 4.80925792 3.556416667
4 3/19/17 winter grandis 10.8 4.80925792 3.556416667
4 3/19/17 winter cockerelli 10.9 4.936758182 3.556416667
1 3/19/17 winter cockerelli 11 5.066427825 3.556416667
2 3/19/17 winter cockerelli 11 5.066427825 3.556416667
1 3/19/17 winter grandis 11 5.066427825 3.556416667
1 3/19/17 winter cockerelli 11.2 5.332341968 3.556416667
2 3/19/17 winter cockerelli 11.5 5.747899893 3.556416667
这是我的代码：
newdata = split(data, list(data$Species, data$Site))
seasonCompare = lapply(newdata, function(x) {
m.interaction = (lm(BodySize ~ SeasonGDD*Site, data = x))
#Anova table
anova(m.interaction)
# Obtain slopes
m.interaction$coefficients
m.lst = lstrends(m.interaction, "Site", var="SeasonGDD")
# Compare slopes
print(pairs(m.lst))
})
这是我希望输出看起来像的样子：
[[1]]$cockerelli.1
P value adjustment: tukey method for comparing a family of 5 estimates
contrast estimate SE df t.ratio p.value
1 - 2 0.0044905951 0.0020291812 525 2.213 0.1764
1 - 3 0.0007809249 0.0004559391 525 1.713 0.4269
1 - 4 0.0022220434 0.0004862632 525 4.570 0.0001
1 - 5 0.0610176389 0.0141858874 525 4.301 0.0002
2 - 3 -0.0037096702 0.0020046493 525 -1.851 0.3457
[[1]]$infrequens.1
P value adjustment: tukey method for comparing a family of 5 estimates
contrast estimate SE df t.ratio p.value
1 - 2 -4.003309e-04 0.0002637222 3067 -1.518 0.5508
1 - 3 -2.272248e-04 0.0001984400 3067 -1.145 0.7825
1 - 4 -3.664736e-04 0.0001983656 3067 -1.847 0.3464
非常感谢！
不需要的输出：
contrast estimate SE df t.ratio p.value
1 - 2 -0.010726867 0.0014693633 345 -7.300 <.0001
1 - 3 -0.008197681 0.0009939915 345 -8.247 <.0001
1 - 4 -0.009577345 0.0010252581 345 -9.341 <.0001
1 - 5 -0.424121019 0.0759824627 345 -5.582 <.0001
2 - 3 0.002529186 0.0013368246 345 1.892 0.3236
2 - 4 0.001149522 0.0013602334 345 0.845 0.9163
2 - 5 -0.413394152 0.0759877208 345 -5.440 <.0001
3 - 4 -0.001379664 0.0008241514 345 -1.674 0.4514
3 - 5 -0.415923338 0.0759800152 345 -5.474 <.0001
4 - 5 -0.414543674 0.0759804306 345 -5.456 <.0001
P value adjustment: tukey method for comparing a family of 5 estimates
contrast estimate SE df t.ratio p.value
1 - 2 -3.445648e-06 0.0008580996 279 -0.004 1.0000
1 - 3 3.055790e-03 0.0011595803 279 2.635 0.0240
2 - 3 3.059235e-03 0.0012805524 279 2.389 0.0461
答案 0 :(得分：0)
请勿使用lapply，而使用sapply-它会保留列名。
请查看sapply输出的效果如何：
> df <- data.frame(one=c(1,1), foo=c(4,5))
> lapply(df, mean)
$one
[1] 1
$foo
[1] 4.5
> sapply(df, mean)
one fuzz
1.0 4.5
答案 1 :(得分：-1)
尝试：
names(seasoneCompare) <- names(newdata)